The angles of the triangle ABC are proportional to the numbers 2: 5: 8. Find the outside and inside corners of a triangle ABC.
Let the smaller inner angle of the triangle be 2 * X0, then the average, in terms of magnitude, the angle will be 5 * X0, and the larger angle will be 8 * X0.
The sum of the interior angles of a triangle is 180, then:
2 * X + 5 * X + 8 * X = 180.
15 * X = 180.
X = 180/15 = 12.
Then the angle ACB = 2 * 12 = 240, angle CAB = 5 * 12 = 60, angle ABC = 8 * 12 = 96.
Since the outer corner of the triangle is an adjacent angle with the inner corner of the triangle at this vertex, and their sum is 180, then the angle AСM = 180 – 24 = 156, the angle СBK = 180 – 96 = 84, the angle BAP = 180 – 60 = 120.
Answer: The inner angles of the triangle are: 24, 60, 96. The outer angles of the triangle are: 156, 120, 84.
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